Problem 39 - Calculus Single Variable

Question

Solve the given differential equation. $$ \exp (x+y) \frac{d y}{d x}=x $$

Step-by-Step Solution

Answer

The solution is \( y = \ln(-\exp(-x)(x + 1) + C) \), where \( C \) is a constant.

Step 1: Simplify the Equation

We start with the given differential equation: \( \exp(x + y) \frac{dy}{dx} = x \). The exponential term \( \exp(x+y) \) makes it difficult to solve directly, so let us rewrite it by separating variables. First, divide both sides of the equation by \( \exp(x+y) \): \[ \frac{dy}{dx} = \frac{x}{\exp(x+y)}. \]

Step 2: Separate the Variables

We aim to separate the variables so that all instances of \( y \) are on one side of the equation and \( x \) on the other. Rewrite the equation as follows:\[ \frac{dy}{dx} = x \cdot \exp(-x) \cdot \exp(-y) \]\[ \exp(y) \, dy = x \exp(-x) \, dx. \]This allows us to separate and integrate both sides:

Step 3: Integrate Both Sides

Integrate each side separately: \[ \int \exp(y) \, dy = \int x \exp(-x) \, dx. \]For the left side, the integration is straightforward:\[ \int \exp(y) \, dy = \exp(y) + C_1. \]For the right side, we use integration by parts. Let \( u = x \) and \( dv = \exp(-x) \, dx \). Then \( du = dx \) and \( v = -\exp(-x) \). This gives:\[ \int x \exp(-x) \, dx = -x \exp(-x) + \int \exp(-x) \, dx = -x \exp(-x) - \exp(-x) + C_2. \]

Step 4: Combine and Simplify

Now, combine the integrations:\[ \exp(y) + C_1 = -x \exp(-x) - \exp(-x) + C_2. \]Let \( C = C_2 - C_1 \), combining constants onto one side:\[ \exp(y) = -x \exp(-x) - \exp(-x) + C. \]Simplify the right side:\[ \exp(y) = -\exp(-x)(x + 1) + C. \]

Step 5: Solve for y

The expression \( \exp(y) = -\exp(-x)(x + 1) + C \) suggests a solution for \( y \) as:\[ y = \ln(-\exp(-x)(x + 1) + C). \]This provides the implicit solution to the differential equation.

Key Concepts

Variable Separation

Variable separation is a powerful technique for solving differential equations, especially those that can be expressed in a form where all terms involving one variable are on one side of the equation and the terms involving the other variable are on the opposite side. Let's understand this concept more clearly with the example at hand.

The differential equation we are considering is \( \exp(x+y) \frac{dy}{dx} = x \). Our goal is to manipulate this equation in such a way that we have all \( y \) terms on one side and all \( x \) terms on the other side, facilitating the integration process.

By dividing both sides by \( \exp(x+y) \) and rearranging, we separate the variables as:

• Move \( \exp(y) \) to the \( dy \) side: \( \exp(y) \, dy \)
• Move \( \exp(-x) \) and \( x \) to the \( dx \) side: \( x \exp(-x) \, dx \)

This results in: \( \exp(y) \, dy = x \exp(-x) \, dx \).

Once separated, we can proceed with integrating each side independently. This step is crucial because it simplifies the problem, making it solvable using standard integration techniques.

Integration by Parts

Integration by parts is an essential tool, especially when dealing with products of functions, like in our differential equation. The formula used is derived from the product rule of differentiation and can be written as:

\[ \int u \, dv = uv - \int v \, du \]

For the equation \( \int x \exp(-x) \, dx \), we choose:

• \( u = x \), so \( du = dx \)
• \( dv = \exp(-x) \, dx \), which gives \( v = -\exp(-x) \)

This setup leads to:

\[ \int x \exp(-x) \, dx = -x \exp(-x) + \int \exp(-x) \, dx \]

We then integrate \( \exp(-x) \, dx \) to find:

\[ \int \exp(-x) \, dx = -\exp(-x) \]

Thus, the full integration yields:

\[ \int x \exp(-x) \, dx = -x \exp(-x) - \exp(-x) + C \]

Integration by parts is an invaluable technique when variables are intertwined multiplicatively, and it lays the groundwork for finding solutions to complex differential equations.

Implicit Solution

An implicit solution of a differential equation expresses the relationship between variables without explicitly solving for one variable in terms of others. In this scenario, the implicit solution comes into play after integrating both sides of the separated equation and consolidating them.

We started with:

\[ \exp(y) + C_1 = -x \exp(-x) - \exp(-x) + C_2 \]

By combining constants into a single constant \( C \), we simplify the equation to:

\[ \exp(y) = -\exp(-x)(x + 1) + C \]

This expression gives the implicit solution because \( y \) cannot be fully isolated due to the natural log function on one side. Yet, it captures the essence of the solution to the differential equation, showing a functional relationship between \( x \) and \( y \), which can be further solved for \( y \) if specific values or initial conditions are provided.

This type of solution is typical in nonlinear differential equations, highlighting the interconnectedness of the involved variables.

Exponential Function

The exponential function \( \exp(x) \) or \( e^x \) is a crucial concept in solving differential equations and is characterized by its constant growth rate, where the rate of change is proportional to its current value.

In our differential equation \( \exp(x+y) \frac{dy}{dx} = x \), the exponential function plays a pivotal role in both the separation of variables and the integration process.

The strategy begins with transforming terms like \( \exp(x+y) \) into separate factors, such as \( \exp(x) \exp(y) \), allowing for manipulation and simplification of the equation. Once separated, integrating functions like \( \exp(y) \) is straightforward, yielding:

\[ \int \exp(y) \, dy = \exp(y) + C \]

This straightforward behavior is one reason the exponential function is so valuable in calculus and differential equations. Its derivatives and integrals have simple expressions, aiding in both simplification and solution of complex differential equations.

Understanding how to handle exponential functions—their creation and cancellation—is essential for mastering differential calculus.